Optical instruments are how we see the world, from corrective eyewear to medical endoscopes to cell phone cameras to orbiting telescopes. This course will teach you how to design such optical systems with simple mathematical and graphical techniques. The first order optical system design covered in the previous course is useful for the initial design of an optical imaging system but does not predict the energy and resolution of the system. This course discusses the propagation of intensity for Gaussian beams and incoherent sources. It also introduces the mathematical background required to design an optical system with the required field of view and resolution. You will also learn how to analyze these characteristics of your optical system using an industry-standard design tool, OpticStudio by Zemax.

Avis

YS

The content is good, and the instructor is very responsive through email. Though I think the capstone is not as challenging as the first course.

OD

Nov 15, 2018

Filled StarFilled StarFilled StarFilled StarFilled Star

The lab demonstrations were very helpful and the explanations of complex phenomena were very easy to understand.

À partir de la leçon

Radiometry

One of the main questions you ask when designing an optical system is "How much light can I get through the system?" In this last section of new content for this course, we move from talking about resolution to talking about the amount of light we expect at each point in the optical system, a field of study called radiometry.

Enseigné par

Amy Sullivan

Research Associate

Robert McLeod

Professor

Transcription

Now, let's start doing some of the math. I'm going to do several different examples here, none of them are more fundamental than the other, but I want you to simply see how we do the simple trigonometry keeping track of these units carefully to understand common questions about light sources and how they'd interact with our optical systems. Let's start up with a simple one which is the point source. And let's ask a question that, if I have a point source, what irradiance, power per unit area, would I expect on a plane? So, if this was a piece of film, or a camera chip, what would the film receive? Or the camera chip on the council they see and access and how would that fall off as I went away from the axis. And what you're going to see here just have to be very careful with cosines, because it's all cosines and sometimes they're in the numerator and sometimes the denominators. So I want to give you these examples. So, let's take our definition of irradiance. What we were not carefully calling intensity before, but now we're not going to use that phrase. It says in watts per unit area and it describes the power per unit area on a plane. And we'd like to relate that to the intensity watts per steradian of the point source at a particular distance away. So, we're going to have some power, we're going figure out this is falling on the plane, and it's going to fall in some little patch of area A here. So I've shown normal here in red to keep it distinct from the tilted case. And so, we're a distance along the normal or away. Well, the first thing we can notice is as we tilt, the area that the point source illuminates goes up. If we think about the same little patch of area A in the untilted case, we see that it's projection onto the plane gets bigger like one over cosine theta. So the infinitesimal patch of area we're illuminating is like A, some finite patch normal incidents, over cosine theta. All right. The next thing we can do is, what's the total power? Well, that's easy. If we're radiating into a patch A, then we can just use the definition of intensity and say that the power is the intensity over the solid angle, which we'll figure out next, corresponding to that area A. And so, by definition, power is intensity times solid angle. Now, let's figure what this solid angle is. Well, we figured out the relationship just a bit ago or actually hit it just by definition, the solid angle, omega, is simply the area over the radius of the sphere that defines that cone. So on axis, that would be A over R squared. However, as we tilt up, we see that we still have the same A, because that's the normal to my radius vector here, but the distance, the radius of the corresponding sphere goes up. In other words, as we tilt up, the solid angle that we're accepting off the point source goes down because the radius is going up. So, over all we find that the irradiance power per unit area on a plane seems to fall like cosine cubed. One cosine comes from the projection of the unit area and two cosines come from the radius squared going up. So, falls off pretty quickly is the point. And if you were writing animation program to make accurate looking scenes of the real world, you probably wouldn't have guessed that you need to make light sources, candles, outside next to a wall that the illumination of that wall falls off really quite quickly. Now let's do an extended source. That was the point source, now let's look at an extended source. And we're going to find that this is one of the places I find that students often get confused that the cosine theta here shows up upside down from the example we just did. So I've tried to color code things and make it real clear, but make sure you understand why the two seem different. So, we think about extended sources as a bunch of point sources. Because they're point sources, they have some intensity, they radiate out into a solid cone, but because there's a bunch of them, they also have an area. So now we need to think about both the angle that our extended source radiates into and it's area. A really good example of this is probably the computer or tablet or phone screen that you are looking at right now. Every point on that screen is a little point radiator. It's radiating out with some cone of angles that determines how easily you can tilt the screen and still see your image or my image and, of course, it also has an area. So, clearly, these are the two important concepts and once again this should sound really familiar. We're asking questions about area and angle simultaneously. So this is why we describe light sources through their radiance, their power per area per solid angle. So, first, let's assume that all the little point sources on your computer screen, lets say, are isotropic. They radiate like a filament, like a black body, into all angles with equal amounts of power is independent of what angle they are radiating into. So if that was the case, then we could calculate what the radiance L is as a function of what angle we're looking at off of your screen. So this would be like you're tilting your screen or your piece of paper. And that's relatively easy. We again just start with the definition of power per solid angle per unit area. And now here's where that cosine theta turns upside down because now the projection is backwards. Now as we look at the area on your computer screen, maybe this is one pixel of an area A, and we tilt the computer screen and look at it sideways, the apparent area that we see gets smaller by cosine theta. And go back and look at the point source case and see how this is different because this is the kind of place we make mistakes. So, what that would mean is that the apparent radiance of the source would go up like a cosine theta in comparison to the axis. And so, this is the example I had to do with the piece of paper is you should have if the paper was radiating isotropically when you line the paper up so it was pointing right at your eye, all of the isotropic radiators on there should have added up and all simultaneously sent a lot of power towards your eyeball and much less when you were looking at them face on as opposed to edge on. And you didn't see that. And the reason was codified by a guy named Lambert. And this is essentially a phenomenological law where he observed your piece of paper and he said it doesn't seem to go up and, as a matter of fact, if you do some more careful measurements you find in general the radiance doesn't depend for most objects on the angle that you look at. If this was a passenger or a power meter over here. So he said, "Well, it must be the case then that we don't have isotropic point sources in most extended sources. Instead, the intensity of the point sources must fall off like the cosine of the angle that's the normal to the surface." Why? He didn't know. Turns out if you go look at black bodies you can actually figure this out. We have the physics now. But he simply compensated for this cosine theta here. So this is Lambert's law. And we've already used this language. Most extended sources are referred to as Lambertian sources. And that gets this law down here. This is the really important one to remember. If we now instead of putting in an isotropic point source here, that the power doesn't depend on angle, instead we put this point source where the power depends on the cosine of the angle, the cosine in the intensity of the point sources compensates for the cosine that we get from the area being projected. And we find that the radiance of most extended sources doesn't depend on angle, it's a constant. And that's what you see when you tilt your piece of paper. So this is something to know about imaging off of most surfaces in the world, for example, is that their radiance is independent of angle. And that's got some pretty big implications for then how you'll find power getting off of those and into your optical system.